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Điều kiện xác định của biểu thức \(A\)là: 

\(\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\\x-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\).

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)

Khi \(x=9\): \(A=\frac{\sqrt{9}}{\sqrt{9}+1}=\frac{3}{4}\).

\(B=A\left(x-1\right)=\frac{\sqrt{x}}{\sqrt{x}+1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\)

\(=x-\sqrt{x}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu \(=\)khi \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}\).

ĐK: \(x\ge0,x\ne1\).

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right)\div\frac{1}{\sqrt{x}+1}\)

\(=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\)

\(=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\left(\sqrt{x}+1\right)\)

\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(P=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)(\(x\ge0,x\ne1\)) 

\(\Rightarrow4\left(\sqrt{x}+2\right)=5\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow\sqrt{x}=13\)

\(\Leftrightarrow x=169\)(tm) 

\(M=\frac{x+12}{\sqrt{x}-1}.\frac{1}{P}=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}=\frac{x-4}{\sqrt{x}+2}+\frac{16}{\sqrt{x}+2}\)

\(=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\sqrt{x}+2+\frac{16}{\sqrt{x}+2}-4\ge2\sqrt{\left(\sqrt{x}+2\right)\frac{16}{\sqrt{x}+2}}-4=4\)

Dấu \(=\)khi \(\sqrt{x}+2=\frac{16}{\sqrt{x}+2}\Leftrightarrow x=4\)(tm) 

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theo mình là vậy.

ĐK: \(x\ge0,x\ne1\).

\(A=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+1}{\sqrt{x}-1}-\left(\sqrt{x}-1\right)\)

\(=\frac{x-\sqrt{x}+1-\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\)

\(=\frac{x-\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}}{\sqrt{x}-1}\)

Với \(x=\frac{9}{4}\): 

\(A=\frac{\sqrt{\frac{9}{4}}}{\sqrt{\frac{9}{4}}-1}=\frac{\frac{3}{2}}{\frac{3}{2}-1}=3\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-1}< 1\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-1}-1< 0\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow0\le x< 1\).

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Weo
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\(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(=\frac{\left(x\sqrt{y}-y\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{\sqrt{xy}.x+xy-xy-\sqrt{xy}.y}{x-y}\)

\(=\frac{\sqrt{xy}.x-\sqrt{xy}.y}{x-y}\)

\(\frac{=\sqrt{xy}.\left(x-y\right)}{x-y}\)

\(=\sqrt{xy}\)

a, \(\sqrt{16x}=8\)ĐK : x >= 0 

\(\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)ĐK : x >= 1 

\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)